Identity of $8\sin^2(t)\cos^2(t)$

Question

I know this probably has a simple answer, but I am having trouble understanding the steps to find the identity for this problem.

This is the answer I was provided:

$$8\sin^2(x)\cos^2(x) = 2\sin^2(2x)$$

The closest Identity I can find is:

$$\sin(x)\cos(y) = 1/2[\sin(x+y) + \sin(x-y)]$$

Which would give

$$\frac 1 2 [\sin(x+x) + \sin(x-x)] = \frac 1 2 [\sin(2x)+\sin(0)] = \frac 1 2 [\sin(2x)+1]$$

Which plugged in would give: $4(\sin(2x) + 1)$

Which clearly isn't what I'm looking for... tell me what I'm doing wrong please!

You've made a mistake. sin(0) = 0, it's cos(0) = 1. Then the formua reduces to sin(x)cos(x) = 1/2*sin(2x), which should let you get the desired result — Jason Carr, Dec 03 '15 at 03:17
@JasonCarr Ah! The simple mistakes get me :| that makes sense, thanks! — acupofjose, Dec 03 '15 at 03:18

score 2 · Answer 1 · answered Dec 03 '15 at 03:18

2

HInt: Use $\sin(2x)=2\sin{x}\cos{x}$

answered Dec 03 '15 at 03:18

Kushal Bhuyan

7,110

score 2 · Accepted Answer · answered Dec 03 '15 at 03:19

2

Using the following identity, it's pretty straightforward: $$\sin(2x)=2\sin(x)\cos(x)$$ $$\sin^2(2x)=4\sin^2(x)\cos^2(x)$$ $$2\sin^2(2x)=8\sin^2(x)\cos^2(x)$$

answered Dec 03 '15 at 03:19

omega-stable

1,235

score 2 · Answer 3 · answered Dec 03 '15 at 03:19

For all $x,y \in \Bbb{R}$ we have $$ \sin (x+y) = \sin x \cos y + \cos x \sin y, $$ which can either be proved or be as a starting point. To prove it we need a few axioms; see for example Apostol's calculus.

Now, if $x \in \Bbb{R}$, then applying the theorem above gives $$ \sin (x+x) = \sin 2x = 2\sin x \cos x; $$ and you can continue from here.

score 2 · Answer 4 · answered Dec 03 '15 at 03:24

2

Notice, your mistake, $\color{red}{\sin 0\ne 1}$

using trig identity $2\sin A\cos A=\sin 2A$, $$8\sin^2 x\cos^2 x=2(2\sin x\cos x)^2$$ $$=2(\sin (2x))^2$$$$=2\sin^2 (2x)$$

answered Dec 03 '15 at 03:24

Harish Chandra Rajpoot

37,450

Identity of $8\sin^2(t)\cos^2(t)$

4 Answers4